Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.
Example:
Input: [ 1->4->5, 1->3->4, 2->6 ] Output: 1->1->2->3->4->4->5->6
Seen this question in a real interview before?
思路:分制法;逐渐的合并成一个链表;
首先分为两块数组;左右两部分l1\l2;假设这两部分是合并成的 所以再合并这两个链表就可以了;
1、递归分治:
if(left>=right)return lists[left];
int mid=(right-left)/2+left
ListNode *l1=ListSort(Lists,left,mid);注意这里必须是mid 不能是mid-1;
LIstNode *l2=ListSort(Lists,mid+1,right);
然后在将两个链表合并;
2、然后就是合并:
code:
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode *merge(ListNode *l1,ListNode *l2)
{
if(!l1)return l2;
if(!l2)return l1;
if(l1->val<=l2->val)
{
l1->next=merge(l1->next,l2);
return l1;
}
l2->next=merge(l1,l2->next);
return l2;
}
ListNode *ListSort(vector<ListNode*>& lists,int left,int right )
{
if(left>=right)
return lists[left];
int mid=(right-left)/2+left;
ListNode *l1=ListSort(lists,left,mid);//注意这里必须是mid 不能是mid-1;看二分法的基本相关的东西;
ListNode *l2=ListSort(lists,mid+1,right);
return merge(l1,l2);
}
ListNode* mergeKLists(vector<ListNode*>& lists)
{
int n=lists.size();
if(n==0)return nullptr;
return ListSort(lists,0,n-1);
}
};